Proof by Induction

Date: 4/3/96 at 21:10:35
From: Anonymous
Subject: Mathematical Induction
I was given a proof by my math teacher by the means of
mathemetical induction.
Prove i(nCi) = n2^n-1
I was able to make it true for 1, but when it came time to
assume n=k and K=K+1 I got stuck. I came down to this
conclusion so far:
L.S = (k+1)+2(k+1C2)+3(k+1C3)+...

Date: 6/3/96 at 11:27:30
From: Doctor Mike
Subject: Re: Mathematical Induction
Hello Sylvestr@fox.nstn.ca ,
I think the formula your teacher gave is
SUM i*(nCi) = n*2^(n-1)
where the nCi are the famous numbers (called "n Choose i" or
"binomial coefficient") in the n-th row down from the top of
Pascal's Triangle :
1 <-- adds up to 1 = 2^0
1 1 <-- adds up to 2 = 2^1
1 2 1 <-- adds up to 4 = 2^2
1 3 3 1 <-- adds up to 8 = 2^3
1 4 6 4 1 <-- adds up to 16 = 2^4
1 5 10 10 5 1 <-- adds up to 32 = 2^5
etc.
This is a tough problem that will help you grow mathematically.
There is a simpler formula SUM nCi = 2^n that you should
probably do first. The key to the induction step on this one is
the same as the next-row-rule for Pascal's Triangle, namely,
(n+1)Ci = nCi + nC(i-1). Prove this rule without mathematical
induction. For the triangle, the rule says that a number in the
next row is the sum of the 2 numbers right next to it in the row
above. Here is how the induction step goes for the simpler
formula. I'm writing it down in a way that shows the limits of
"i", which you should watch VERY closely. View this with a font
like Courier (all characters, including blank, have same width).
n+1 n n
SUM (n+1)Ci = 2 + SUM (n+1)Ci = 2 + SUM [nCi + nC(i-1)]
i=0 i=1 i=1
n n n n-1
= 2 + SUM nCi + SUM nC(i-1) = 2 + SUM nCi + SUM nCi
i=1 i=1 i=1 i=0
n n n n n+1
= SUM nCi + SUM nCi = 2 + 2 = 2
i=0 i=0
Once you know the simpler formula, you can prove the other one.
That induction step also uses the next-row-rule for Pascal's
Triangle, and it starts like this.
n+1 n
SUM i*(n+1)Ci = (n+1) + SUM i*(n+1)Ci
i=1 i=1
n
= (n+1) + SUM i*[nCi + nC(i-1)]
i=1
n n
= (n+1) + SUM i*nCi + SUM i*nC(i-1)
i=1 i=1
n-1 n
= (n+1) + n*2 + SUM [(i-1)*nC(i-1) + 1*nC(i-1)]
i=1
n-1 n n
= (n+1) + n*2 + SUM (i-1)*nC(i-1) + SUM nC(i-1) =...
i=2 i=1
and it continues similarly to the end of the other proof.
I hope this helps.
-Doctor Mike, The Math Forum